NCERT solutions for Class 12 Physics Chapter 3 - Current Electricity

NCERT Solutions for Class 12 Physics Chapter 3 - Current Electricity: Additional Questions

The Current Electricity NCERT Solutions of Class 12 Physics are meant to be used in addition to the textbook exercises and are aimed at giving an extra practice on the topic to be understood further. Such questions challenge the reasoning, formula application, and clarity of concepts, and the students are exam-ready. With a regular solution, learners will be able to further prepare themselves to not only board exams but also entrance tests such as JEE and NEET.

1. The earth’s surface has a negative surface charge density of 10 -9 C m ^-2 . The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A cover the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10 ⁶ m.)

Answer:

Given,

The surface charge density of Earth

$\rho$ = $10^{-9}C m^{-2}$

Current over the entire globe = 1800 A

Radius of earth, r = 6.37 x 106 m

$\therefore$ The surface area of earth A = $4\pi r^2$

A= $4\pi (6.37\times10^6 )^2$ = $5.09 \times 10^{14} m^2$

Now, charge on the earth's surface,

$q=\rho\times A$

Therefore,

$q=1005.09 \times 10^5 \mathrm{C}$

Let the time taken to neutralize earth surface be t

$\therefore$ Current I = q / t

t = 282.78 s.

Therefore, the time taken to neutralise the Earth's surface is 282.78 s

2. Six lead-acid type secondary cells, each of emf 2.0 V and internal resistance 0.015 Ω, are joined in series to provide a supply to a resistance of 8.5 Ω. What is the current drawn from the supply and its terminal voltage?

Answer:

Given,

There are 6 secondary cells.

Emf of each cell, E = 2 V (In series)

The internal resistance of each cell, r = 0.015 Ω (In series)

And the resistance of the resistor, R = 8.5 Ω

Let I be the current drawn in the circuit.

$I = \frac{nE}{R + nr}$

$\implies I = \frac{6(2)}{8.5 + 6(0.015)} = \frac{12}{8.59}$

$\implies$ I = 1.4 A

Hence current drawn from the supply is 1.4 A

Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V

3. A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

Given,

Emf , E = 1.9 V

Internal resistance, r =380 Ω

The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A

The motor requires a large value of current to start, and hence, this cell cannot be used for a car.

4. Two wires of equal length, one of aluminium and the other of copper, have the same resistance. Which of the two wires is lighter? Hence, explain why aluminium wires are preferred for overhead power cables. ( $\rho_{Al} = 2.63\times 10^{-8} \Omega m$ . $\rho_{Cul} = 1.73\times 10^{-8} \Omega m$ relative density of Al = 2.7, of Cu = 8.9.)

Answer:

We know,

$R = \rho l / A$

The wires have the same resistance and are of the same length.

Hence,

$\frac{\rho_{Al}}{A_{Al}} = \frac{\rho_{Cu}}{A_{Cu}}$

$\implies \frac{A_{Al}}{A_{Cu}} = \frac{\rho_{Al}}{\rho_{Cu}} = \frac{2.63}{1.73}$

Now, mass = Density x Volume = Density x Area x length

Taking the ratio of their masses for the same length

$\implies \frac{m_{Al}}{m_{Cu}} = \frac{d_{AL}A_{Al}}{d_{Cu}A_{Cu}} = \frac{2.7\times2.63}{8.9\times1.73} < 1$

Hence, $m_{Al} < m_{Cu}$

Therefore, for the same resistance and length, the aluminium wire is lighter.

Since aluminium wire is lighter, it is used in power cables.

5. What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Answer:

The ratio of Voltage to current for the various values comes out to be nearly constant, which is around 19.7.

Hence, the resistor made of alloy manganin follows Ohm's law.

6. A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, or drift speed?

Answer:

The current flowing through the conductor is constant for a steady current flow.

Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.

Answer the following questions:

7. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

Answer:

No. Ohm’s law is not universally applicable to all conducting elements.

A semiconductor diode is such an example.

Answer the following questions

8. A low-voltage supply from which one needs high currents must have very low internal resistance. Why?

Answer:

Ohm's law states that: V = I x R

Hence, for a low voltage V, the resistance R must be very low for a high value of current.

Answer the following questions:

9. A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer:

A very high internal resistance is required for a high-tension supply to limit the current drawn for safety purposes.

10. Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with the increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of ( $10^{22} / 10^{23}$ ).

Answer:

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 10²².

11. (a) Given n resistors each of resistance R, how will you combine them to get the maximum effective resistance?

Answer:

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

11. (b) Given n resistors each of resistance R, how will you combine them to get the minimum effective resistance?

Answer:

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

11.(c) What is the ratio of the maximum to the minimum resistance?

Answer:

The ratio is nR/(R/n) = $n^2$

12.(a) Given the resistances of 1 Ω, 2 Ω, and 3 Ω, how will we combine them to get an equivalent resistance of $\frac{11}{3}\Omega$?

Answer:

We have equivalent resistance = 11/3

Let's break this algebraically so that we can represent it in terms of 1, 2 and 3

$\frac{11}{3}=\frac{9+2}{3}=3+\frac{2}{3}=3+\frac{1*2}{1+2}$

This expression is expressed in terms of 1, 2 and 3. Hence, we can make a circuit that consists only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :

12. (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will we combine them to get an equivalent resistance of $\frac{11}{5}\Omega$?

Answer:

Connect a 2 Ω and a 3 Ω resistor in parallel and a 1 Ω resistor in series to it

Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1

R = 11/5 Ω

12.(c) Given the resistances of 1 Ω, 2 Ω, and 3 Ω, how will you combine them to get an equivalent resistance of 6 Ω?

Answer:

1 Ω+2 Ω+3 Ω=6 Ω, so we will combine the resistances in series.

12.(d) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will you combine them to get an equivalent resistance of $\frac{6}{11}\Omega$?

Answer:

Connect all three resistors in parallel.

Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)

R = 6/11 Ω

13.(a) Determine the equivalent resistance of the networks shown in Fig.

Answer:

It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor, and two 2-ohm resistors are also in series, and we have 4 loops.

The equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is

$Equivalent\ R_{loop}=\frac{2*4}{2+4}=\frac{8}{6}=\frac{4}{3}$

Now we have 4 such loops in series,

$Total\ Equivalent\ R_{loop}=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=\frac{16}{3}$

Hence equivalent resistance of the circuit is 16/3 ohms.

13.(b) Determine the equivalent resistance of the networks shown in Fig

Answer:

It can be seen that all 5 resistors are in series, so

Equivalent Resistance = R + R + R + R + R = 5R

Hence, equivalent resistance is 5R.

14. Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

Answer:

First, let us find the equivalent of the infinite network,

let equivalent resistance = R'

From the figure, we can consider the box as a resistance of R.

Now, we can write,

equivalent resistance = R' =[( R')Parallel with (1)] + 1 + 1

$\frac{R'*1}{R'+1}+2=R'$

$R'+2R'+2=R'^2+R'$

$R'^2-2R'-2=0$

$R'=1+\sqrt{3},or1-\sqrt{3}$

Since resistance can never be negative, we accept

$R'=1+\sqrt{3}$

, We have calculated the equivalent resistance of an infinite network,

Now

Total equivalent resistance = internal resistance of the battery + equivalent resistance of the infinite network

= 0.5+1+1.73

=3.23 ohm

$V=IR$

$I=\frac{V}{R}=\frac{12}{3.23}=3.72A$

Hence current drawn from the 12V battery is 3.72 amperes.

15. Figure 3.33 shows a potentiometer with a cell of 2.0 V and an internal resistance of 0.40 Ω, maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value of $\epsilon$?

Answer:

Given

Maintained a constant emf of the standard cell = 1.02V, balanced point of this cell = 67.3cm

Now, when the standard cell is replaced by another cell with emf = $\varepsilon$, the balanced point for this cell = 82.3cm

Now, as we know the relation

$\frac{\varepsilon}{l} =\frac{E}{L}$

$\varepsilon =\frac{E}{L}*l=\frac{1.02}{67.3}*82.3=1.247V$

Hence, the emf of another cell is 1.247V.

16. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(b) What purpose does the high resistance of 600 k $\Omega$ have?

Answer:

If a sufficiently high current passes through the galvanometer, then it can get damaged. So we limit the current by adding a high resistance of 600 k $\Omega$.

17. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(c) Is the balance point affected by this high resistance?

Answer:

No, the Balance point is not affected by high resistance. High resistance limits the current to the galvanometer wire. The balance point is obtained by moving the joystick on the potentiometer wire, and the current through the potentiometer wire is constant. The balance point is the point at which the current through the galvanometer becomes zero. The only duty of the high resistance is to supply a limited, constant current to the potentiometer wire.

18. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

Answer:

No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when the emf of the driving point is less than that of the other cell, there won't be any balance point in the wire.

19. Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit?

Answer:

No, the circuit would not work properly for a very low order of Voltage because the balance points would be near point A, and there would be a greater percentage error in measuring it. If we add a series resistance with wire AB. It will increase the potential difference of wire AB, which will lead to a decrease in percentage error.

20. Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer:

Given,

The balance point of the cell in an open circuit = $l_1=76.3cm$

value of external resistance added = $R=9.5\Omega$

new balance point = $l_2=64.8cm$

Let the internal resistance of the cell be $r$.

Now as we know, in a potentiometer,

$r=\frac{l_1-l_2}{l_2}*R$

$r=\frac{76.3-64.8}{64.8}*9.5=1.68\Omega$

Hence, the internal resistance of the cell will be 1.68 $\Omega$.

Class 12 Physics NCERT Chapter 3: Higher Order Thinking Skills (HOTS) Questions

The Class 12 Physics Chapter 3 HOTS (Higher Order Thinking Skills) questions on Current Electricity are designed to test deep conceptual understanding and analytical thinking. These questions go beyond direct textbook problems, encouraging students to apply concepts like Ohm’s Law, Kirchhoff’s Laws, and electrical circuits in new situations. Solving these helps students strengthen logical reasoning, improve problem-solving accuracy, and prepare effectively for competitive exams like JEE and NEET.

Q.1 The total current supplied to the circuit by the battery is

Answer:

The equivalent circuit is
The total resistance of the circuit is

$1.5+\frac{2 \times 6}{2+6}$ in parallel with $3 \Omega$
$\therefore$ Total resistance is $\frac{3}{2}=1.5 \Omega$
Current in the circuit $=\frac{6}{1.5}=4 \mathrm{~A}$

Q.2 What should be the value of E for which the galvanometer shows no deflection?

Answer:

If there is no current in G, as the potential difference across $10 \Omega$ resistance is equal to 10 V, then

$
\begin{aligned}
& \frac{\frac{10}{15}+\frac{E}{5}}{\frac{1}{15}+\frac{1}{15}}=10 \\
& \Rightarrow E=10 \mathrm{~V}
\end{aligned}
$

Q.3 A circuit consists of five identical conductors as shown in the figure. The two similar conductors are added as indicated by the dotted lines. The ratio of resistances before and after addition will be:

Answer:

Before addition,

Total resistance of the circuit is $\mathrm{R}_1=5 \Omega$
After the addition of the two conductors, the circuit will acquire the form shown in the figure

It is a balanced Wheatstone bridge. The resistance of the central conductor is ineffective.
$\therefore$ Total resistance of the circuit is

$
\begin{aligned}
& \mathrm{R}_2=1 \Omega+\frac{(2 \Omega)(2 \Omega)}{(2 \Omega+2 \Omega)}+1 \Omega=1 \Omega+1 \Omega+1 \Omega=3 \Omega \\
& \therefore \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5}{3}
\end{aligned}
$

Q.4 A current of $2A$ flows in the system of conductors as shown in the figure. The potential difference $V_P-V_R$ will be:

Answer:

Current through QRS = current through $\mathrm{}$ QPS=1$ \mathrm{}$ amp.

$
\begin{aligned}
& V_D-V_A=2 \times 1=2 \text { volt } \\
& V_D-V_B=3 \times 1=3 \text { volt } \\
& \therefore V_A-V_B=\left(V_D-V_B\right)-\left(V_D-V_A\right) \\
& =3-2 \\
& =1 \text { volt }
\end{aligned}
$

Q.5 In the network shown in the figure below, the potential difference between A and B is

Answer:
The distribution of current is shown in Fig. Keeping in view that the inflow and outflow of current in a cell must be the same. Applying the loop rule to left and right loops.


$\begin{aligned} & 2 \mathrm{i}_1+6 \mathrm{i}_1=4 \text { or } 2 \mathrm{i}_1=0.5 \mathrm{~A} \\ & 3 \mathrm{i}_1+1 \mathrm{i}_2=6 \text { or } \mathrm{i}_2=1.5 \mathrm{~A} \\ & \mathrm{~V}_{\mathrm{AB}}=\sum \mathrm{ir}-\sum e \\ & (=-2 \times 0.5+3 \times 1.5-4=-0.5 \mathrm{~V}) \\ & \mathrm{V}_{\mathrm{AB}}=-0.5 \mathrm{~V}\end{aligned}$